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S_A_V [24]
2 years ago
11

Help me please!

Chemistry
2 answers:
Vlad1618 [11]2 years ago
8 0

Answer:

  • C. 108 grams/100g of H2O

Required number is the vertical coordinate of the intersection point of a line at 60°C with the graph of the KNO₃.

Dmitry_Shevchenko [17]2 years ago
7 0

Answer:

Explanation:

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dependent variable I think

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22. What is a comet?
Lunna [17]

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A ball of ice blasted into outer space

Explanation:

It is a ball of ice that is going so fast it leaves a "tail"

4 0
3 years ago
What is the ratio of the areas of the signals in the h nmr spectrum of pentan-3-ol?
Damm [24]

The ratio of the areas of the signals in the h NMR spectrum of pentan-3-ol is 6: 4: 1: 1. The correct option is A.

<h3>What is a NMR spectrum?</h3>

Nuclear magnetic resonance spectroscopy is a spectroscopy that shows the detailed structure and chemical environment of a chemical element.

Pentan-3-ol contain 12 hydrogen atoms. In H-NMR spectra, hydrogen atoms have same environment gives a signal.

There are 4 different kinds of signals due of the 4 different environment experienced by these 12 hydrogens.

Thus, the  ratio of the areas of the signals in the h NMR spectrum of pentan-3-ol is 6: 4: 1: 1. The correct option is A.

Learn more about NMR spectrum

brainly.com/question/9812005

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8 0
2 years ago
Kp for the following reaction is 0.16 at 25 degree C. 2 NOBr(g) 2 NO(g) Br_2(g) The enthalpy change for the reaction at standard
finlep [7]

Answer:

Explanation:

Given that:

2 NOBr_{(g)} \iff 2 NO_{(g)} + Br_{2(g)}

From above:

K_p = 0.16 = \dfrac{(P_{NO})^2 (P_{Br})}{(P_{NOBr})^2}

To predict the effect of the addition of Br₂(g);

The addition of Br₂(g) will favor the equilibrium to shift to the left i.e. formation of NOBr

The removal of some NOBr will cause the equilibrium position to shift to the left side. This is because concentration on the left side is decreased and the concentration on the right side will be increased. Thus, the equilibrium will shift towards where the concentration is reduced which is the left side.

5 0
3 years ago
Consider the reaction: I2(g)+Cl2(g)⇌2ICl(g) Kp= 81.9 at 25 ∘C. Calculate ΔGrxn for the reaction at 25 ∘C under each condition: -
scoray [572]

Answer:

Part 1: - 1.091 x 10⁴ J/mol.

Part 2: - 1.137 x 10⁴ J/mol.

Explanation:

Part 1: At standard conditions:

At standard conditions Kp= 81.9.

∵ ΔGrxn = -RTlnKp

∴ ΔGrxn = - (8.314 J/mol.K)(298.0 K)(ln(81.9)) = - 1.091 x 10⁴ J/mol.

Part 2: PICl = 2.63 atm; PI₂ = 0.324 atm; PCl₂ = 0.217 atm.

For the reaction:

I₂(g) + Cl₂(g) ⇌ 2ICl(g).

Kp = (PICl)²/(PI₂)(PCl₂) = (2.63 atm)²/(0.324 atm)(0.217 atm) = 98.38.

∵ ΔGrxn = -RTlnKp

∴ ΔGrxn = - (8.314 J/mol.K)(298.0 K)(ln(98.38)) = - 1.137 x 10⁴ J/mol.

6 0
3 years ago
Read 2 more answers
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